The Formula for the Finite Sum of Consecutive Positive Integers

Consider the sum $\sum_{j=1}^{n} j = 1 + 2 + ... + n$ for some $n \in \{1, 2, ... \}$. You are likely already familiar with the finite sum of positive integers formula which is given to be:

(1)

\begin{align} \sum_{j=1}^{n} j = \frac{n(n+1)}{2} \end{align}

The result above can be proven rather easily using mathematical induction. It can also be proven geometrically. We will instead look at a combinatorial proof that instead derives the formula above. We must first look at the following theorem.

Proof: Recall from the Sums of Binomial Coefficients page that if $n$ and $k$ are nonnegative integers that satisfy $1 \leq k \leq n$ then $\displaystyle{\sum_{j=1}^{n} \binom{j}{k} = \binom{n+1}{k+1}}$. If we multiply both sides of this equality by $k!$ then we get: